Question

Towns A and B in Figure are $\mathbf{80}\mathbf{}\mathbf{\text{km}}$apart.couple arranges to drive from town A and meet a couple driving from town B at the lake, L. The two couples leave simultaneously and drive for $\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{\text{h}}$in the directions shown. Car 1 has a speed of $\mathbf{90}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{h}$. If the cars arrive simultaneously at the lake, what is the speed of car 2?

Short answer

The speed of car 2 is $68.4\mathrm{km}/\mathrm{h}$.

Step-by-step solution

Given data:

The speed of car 1 is $90.0\mathrm{km}/\mathrm{h}$and moves in a direction at ${40}^{\circ }$from the horizontal. The distance between town A and B is $84.0\mathrm{km}$, the time take by the couples to drive to the lake is$2.5\text{h}$.

Distance formula:

The expression for the distance is given by,

$D=vt$

Here, $D$is the distance travelled, $v$is the speed and $t$is the time taken.

Determine the speed of the car:

The distance travelled by car 1 is,

$\mathrm{AL}=v_{1}t$

Here,

$v_1$is the speed of car 1, $t$is the time taken by the car to reach the lake.

Substitute $90.0\mathrm{km}/\mathrm{h}$for $v_1$and $2.50\text{h}$for$t$in the above equation.

$\begin{array}{c}\mathrm{AL}=(90.0\mathrm{km}/\mathrm{h})(2.50\mathrm{h})\\ =(90.0\mathrm{km}/\mathrm{h})(2.50\mathrm{h})\\ =225\mathrm{km}\end{array}$

Thus, the distance travelled by car 1 is $225\text{km}$.

From the cosine law in the given figure,

$(\mathrm{BL}{)}^2=(\mathrm{AL}{)}^2+(\mathrm{AB}{)}^2-2\left(\mathrm{AL}\right)\left(\mathrm{AB}\right)\cos \left(40.0\right)$

Substitute $80\text{km}$for $AB$and $225\text{km}$for $AL$in the above equation.

$\begin{array}{c}(\mathrm{BL}{)}^2=(225\mathrm{km}{)}^2+(80\mathrm{km}{)}^2-2\left(225\mathrm{km}\right)\left(80\mathrm{km}\right)\cos \left(40.0°\right)\\ (\mathrm{BL}{)}^2=29447\\ \mathrm{BL}=\sqrt{29447}\\ =171 \mathrm{km}\end{array}$

Thus, the distance travelled by car 2 is role="math" localid="1663615625328" $171\text{km}$.

The distance travelled by car 2 is,

$\mathrm{BL}=v_{2}t$

Here,$v_2$ is the speed of car 2.

Substitute role="math" localid="1663615712149" $171\text{km}$for BL and $2.50\text{h}$for$t$ in the above equation.

$\begin{array}{c}171\mathrm{km}=v_2(2.50\mathrm{h})\\ v_2=\frac{171\mathrm{km}}{2.50\mathrm{h}}\\ =68.4\mathrm{km}/\mathrm{h}\end{array}$

Therefore, the speed of car 2 is $68.4\mathrm{km}/\mathrm{h}$.